Input for ARfit ? (EEG analysis)

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Annaëlle 2024-11-5，15:43

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评论： William Rose 2024-11-12，14:26

Hello,

I am trying to use the ARfit toolbox on EEG data, more specifically the arfit function itself, but I can't figure out what format the input should be in.

It says on the toolbox page that "The input matrix v must contain the time series data, with columns of v representing variables and rows of v representing observations" but I don't understand what it means. What I have is a vector where the rows are the EEG channels, and the columns are the signal values for each data point. However, it clearly doesn't work (either returns an error "time series too short", or it runs but the output is wrong (siglev = 0 with the function arres)).

I'm hoping someone there know how to use the toolbox better than I do... Thanks in advance.

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Answer 1

William Rose 2024-11-5，16:06

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编辑：William Rose 2024-11-5，16:08

在 MATLAB Online 中打开

@Annaëlle,

[Edit: fix spelling.]

Have you run the script ardem.m, which is part of the ARfit toolbox? Does it produce reasonable output?

Assuming it does, I would modify ardem, so that it uses two columns of your data, instead of using the simulated data that it does use. To do this, change line

v = arsim(w, A, C, 200); % generate simulated data v (200x2 array)

to something like

myData=load('myData.txt');  % read data from file
v=myData(1:200,1:2);        % v=rows 1-200 & columns 1 and 2 of data

The details of the new lines above will vary depending on which columns you want to use, etc. But you get the idea. See what happens.

You can also start a discussion on the FileExchange site for that toolbox.

Attach a sample data file here, if you want more assistance.

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Annaëlle 2024-11-5，16:23

No issue running ardem.m.

However, it clearly doesn't work (either returns an error "time series too short", or if it runs but the output is wrong (siglev = 0 with the function arres)). -> This is what happens when I load my data into v, hence why I'm asking for help.

Would it be possible for you to clarrify what the rows and the columns in your data array correspond to?

William Rose 2024-11-5，22:08

@Annaëlle, see @Walter Roberson's answer.

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Answer 2

Walter Roberson 2024-11-5，19:40

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What I have is a vector where the rows are the EEG channels, and the columns are the signal values for each data point.

Your input is in the wrong form for the toolbox.

Each EEG channel should be considered independent. Going to the next row of the data does not correspond to going to the next time-point of the data.

Potentially if you were to flip the array it just might be the right order... provided that what is currently the colums that are signal values for each data point, happen to reflect different times.

You need

channel1_time1 channel2_time1 channel3_time1

channel1_time2 channel2_time2 channel3_time2

channel1_time3 channel2_time3 channel3_time3

channel1_time4 channel2_time4 channel3_time4

and so on.

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William Rose 2024-11-6，17:07

@Annaëlle,

What @Walter Roberson means is that each row of your data array is a different time, and each column is a different channel, or vice versa. If it does not work before you transpose the data array, and it doesn't work after you transpose it, then that's weird.

Please upload a file of the data you want to analyze.

The data array passed to arfit() should have many rows (i.e. measurements at many times) and several columns (i.e. several channels). Here is an example of using arfit().

First, simulate a multiple autoregressive (multi AR) process with 2 channels and 500 time ponts. This could be done with arsim() in the toolbox, but I do it here without the toolbox routine:

>> N=500; % desired simulation duration (points)

>> v=zeros(N+100,2); % allocate v with 100 extra values for simulation startup

>> v(1:2,:)=randn(2,2); % initial values

>> A1 = [0.4 1.2; 0.3 0.7]; % multi AR coeffs for lag=1

>> A2 = [0.35 -0.3; -0.4 -0.5]; % multi AR coeffs for lag=2

>> C=[1 0.5; 0.5 1.5]; % covariance matrix for random noise

>> for i=3:N+100, v(i,:)=(A1*v(i-1,:)'+A2*v(i-2,:)')'+randn(1,2)*chol(C); end

>> v=v(101:end,:); % discard startup (first 100 points)

>> disp(size(v)) % display data array size

500 2

>>

Analyze the data with arfit():

>> [west, Aest, Cest, sbc]=arfit(v,1,5)

west =

-0.0875

-0.0574

Aest =

0.4224 1.1762 0.3259 -0.2827

0.2989 0.6430 -0.4120 -0.4466

Cest =

0.9287 0.4828

0.4828 1.5717

sbc =

0.5327 0.1546 0.1761 0.1987 0.2200

>>

See if you get similar results. (Results won't match exactly, due to randomness.) Then replace the simulated v above with your data array, with successive time points on successive rows, and a column for each channel.

William Rose 2024-11-7，15:49

编辑：William Rose 2024-11-7，15:53

在 MATLAB Online 中打开

v.mat

@Annaëlle,

Let's look at the actual data:

load v

n=length(v);

plot(1:n,v(:,1),'-r',1:n,v(:,2),'-b')

grid on; xlabel('Time (samples)'); ylabel('Amplitude');

legend('Ch.1','Ch.2')

Run arfit() on your data. Specify minimum and maximum lags to be 1 and 5, for now. Screenshot of results is below.

>> [w, A, Cest, sbc]=arfit(v,1,5)

w =

1.0e-10 *

-0.1137

-0.1830

A =

Columns 1 through 8

4.6171 0.0131 -8.8420 -0.0523 8.7743 0.0802 -4.5114 -0.0561

-0.0263 4.6635 0.0803 -8.9806 -0.0903 8.9246 0.0418 -4.5761

Columns 9 through 10

0.9621 0.0150

-0.0054 0.9685

Cest =

1.0e-19 *

0.6669 0.3503

0.3503 0.3031

sbc =

-35.3360 -37.2472 -40.0320 -42.3298 -44.9596

The results look reasonable. A is 2-by-10. arfit chooses the optimum model order (optOrder), within the range minlags to maxlags. The optimum order is the order where sbc is lowest. sbc is smallest at 5 lags. That is why A is 2x10. The size of A is numChan rows by (numChan x optOrder) columns. I cannot explain why you got such a large A matrix.

By the way, if you examine the A matrix, you see that each channel is predicted mainly by itself, and not by the other channel. I say that because A consists of 5 2x2 matrices: one 2x2 matrix for each lag. And the individual 2x2 matrices are almost diagonal. From the results above, we have

A=

4.6171 0.0131 -8.8420 -0.0523 8.7743 0.0802 -4.5114 -0.0561 0.9621 0.0150

-0.0263 4.6635 0.0803 -8.9806 -0.0903 8.9246 0.0418 -4.5761 -0.0054 0.9685

Therefore

A1=[4.62 0.01; -0.03 4.66] (lag 1)

A2=[-8.84 -0.05; 0.08 -8.98] (lag 2)

A3=[8.77 0.08; -0.09 8.92] (lag 3)

etc

Note how A1, A2, A3 are almost diagonal. The arfit package includes routines to compute uncertainty esitmates for all the estimated parametrers. See the package documentation, and ardem() for details. It would not surprise me if the off-diagonal values in the matrices above are not significantly different from 0. If that is the case, then a multiple autoregressive model is not justified for this data, and you should just use individual autoregressive models for each channel independently.

Since sbc was lowest at the maximum order investigated (i.e. at order 5), you could try even higher ranges for the order, to see if you can find a lower sbc (which means a significantly better fit) at a higher order.

Annaëlle 2024-11-12，9:13

Thank you so much for the detailed explanation, it really helped!

William Rose 2024-11-12，14:26

@Annaëlle, you're welcome.

Good luck with your research.

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Input for ARfit ? (EEG analysis)

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